If we let \(\ell _1\)  be the number of rows of \({\mathbf {H}}_1\) , \(\ell _2\)  be the number of rows of the matrix of \({\mathbf {H}}_2^{\prime } \mathop {=}\limits ^{\triangle }\begin{pmatrix} {\mathbf {H}}_1 \\ {\mathbf {H}}_2 \end{pmatrix}\) , then the fact that the elements of \({{L}}^2\)  should have a unique representation in terms of a sum of a pair of elements of \({{L}}^1\)  respectively and that they are all elements \({\mathbf {x}}\)  of weight \(p_1\)  and \(p_2\)  respectively which satisfy \({\mathbf {H}}_2^{\prime } {\mathbf {x}}^\intercal =\mathbf {0}\)  and \({\mathbf {H}}{\mathbf {x}}^\intercal =\mathbf {0}\)  respectively, imposes conditions (REF ) which follow.
The \(S_i\)  represent the space complexity of the successive lists (i.e. \({{L}}^0\) , \({{L}}^1\) , \({{L}}^2\)  and \({{L}}^3\) ) used in the [1]} algorithm, whereas the \(T_i\) 's denote the complexity of each merge and \(\text{T}_{\text{eq}}\)  is the final complexity.
\(2^{\ell _1} &= \binom{p_2}{p_2/2}\binom{n-k+\ell -p_2}{p_1-p_2/2},& 2^{\ell _2} &= \binom{p_3}{p_3/2}\binom{n-k+\ell -p_3}{p_2-p_3/2} \) 
\(S_0 = \binom{\frac{n-k+\ell }{2}}{\frac{p_1}{2}}, \quad S_1 = \frac{\binom{n-k+\ell }{p_1}}{2^{\ell _1}}, \quad S_2 = \frac{\binom{n-k+\ell }{p_2}}{2^{\ell _2}}, \quad S_3 = \frac{\binom{n-k+\ell }{p_3}}{2^{\ell }}\) 
\(T_0 = S_0, \quad T_1 = S_0 + \frac{S_0^2}{2^{\ell _1}}, \quad T_2 = S_1 + \frac{S_1^2}{2^{\ell _2-\ell _1}}, \quad T_3 = S_2 + \frac{S_2^2}{2^{\ell -\ell _2}},\quad \text{T}_{\text{eq}}= T_0+T_1+T_2+T_3\) 
